Worked Examples and Practice

This final lesson brings together everything you have learned across the FSCE 11+ Mathematics course. Below you will find 10 complete exam-style problems, each with a full step-by-step solution and examiner commentary explaining what the examiner is looking for and how marks would be awarded. After the worked examples, you will find a pre-exam checklist and time management advice.

How to Use This Lesson

• Try each problem yourself first before reading the solution.
• Time yourself: aim for about 3-4 minutes per problem.
• After checking the solution, read the examiner commentary — it tells you what earns marks and what loses them.

Problem 1: Number and Place Value

Question: The population of a city is 2,847,193. The mayor says, "Our population is approximately 3 million." Is the mayor correct? Explain your reasoning.

Full solution:

1. To check the mayor's claim, we round 2,847,193 to the nearest million.
2. The millions digit is 2. The digit to the right (hundred thousands) is 8.
3. Since 8 ≥ 5, we round up: 2,847,193 rounds to 3,000,000.
4. 3,000,000 = 3 million, so the mayor is correct.

Examiner commentary: Full marks require both the rounding AND the explanation. Simply saying "yes" earns no marks. Stating the rounding rule ("8 is 5 or more, so round up") earns the reasoning mark. This is a classic FSCE question — they want to see your mathematical thinking, not just a yes/no answer.

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Problem 2: BODMAS

Question: Calculate 120 - 4 x (3 + 7)² ÷ 20

Full solution:

1. Brackets first: 3 + 7 = 10
2. Orders (powers): 10² = 100
3. The expression is now: 120 - 4 x 100 ÷ 20
4. Multiplication and division, left to right: 4 x 100 = 400, then 400 ÷ 20 = 20
5. Subtraction: 120 - 20 = 100

Examiner commentary: The examiner looks for BODMAS applied correctly. The most common error is doing 120 - 4 = 116 first (ignoring BODMAS). Rewriting the expression after each step shows clear method and earns full marks even if there is a small arithmetic slip.

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Problem 3: Fractions

Question: In a school of 420 pupils, 2/7 play football, 1/3 play netball, and the rest play neither. How many pupils play neither sport? Assume no pupil plays both.

Full solution:

1. Football players: 2/7 of 420 = 420 ÷ 7 x 2 = 60 x 2 = 120
2. Netball players: 1/3 of 420 = 420 ÷ 3 = 140
3. Total playing a sport: 120 + 140 = 260
4. Playing neither: 420 - 260 = 160

Examiner commentary: This is a three-step problem. Many pupils get the fraction calculations right but forget to subtract from the total. Read the question again after calculating — it asks for "neither," not "football" or "netball."

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Problem 4: Ratio

Question: A smoothie recipe uses banana, strawberry, and yoghurt in the ratio 3:2:5. If 150g of strawberry is used, how much smoothie is made in total?

Full solution:

1. Strawberry corresponds to 2 parts in the ratio.
2. 2 parts = 150g, so 1 part = 150 ÷ 2 = 75g.
3. Total parts = 3 + 2 + 5 = 10 parts.
4. Total smoothie = 10 x 75 = 750g

Check: Banana = 3 x 75 = 225g. Strawberry = 150g. Yoghurt = 5 x 75 = 375g. Total = 225 + 150 + 375 = 750g. Correct!

Examiner commentary: The key insight is recognising that 150g corresponds to the "2" in the ratio, not the "3" or the total. Finding the value of one part first is the most reliable method. The check (showing all parts add to the total) would impress the examiner.

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Problem 5: Algebra

Question: The nth term of a sequence is 3n + 7. Is 50 a term in this sequence? Explain how you know.

Full solution:

1. If 50 is a term, then 3n + 7 = 50 for some whole number n.
2. Solve: 3n = 50 - 7 = 43
3. n = 43 ÷ 3 = 14.333...
4. Since n is not a whole number, 50 is NOT a term in this sequence.

Reasoning: The position number (n) must be a positive whole number (1st term, 2nd term, 3rd term, etc.). Since 43/3 is not a whole number, there is no position where the sequence equals 50.

Examiner commentary: This is a reasoning question. The mathematical calculation alone (n = 14.33) is not enough — you must explain WHY this means 50 is not in the sequence. State clearly that n must be a whole number. This is exactly the kind of question the FSCE uses to test understanding.

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Problem 6: Geometry

Question: A regular pentagon and a regular hexagon share one side. Find the angle marked x, which is the angle between the unshared sides of the pentagon and hexagon at the shared vertices.

Full solution:

1. Interior angle of a regular pentagon = (5-2) x 180 ÷ 5 = 540 ÷ 5 = 108°
2. Interior angle of a regular hexagon = (6-2) x 180 ÷ 6 = 720 ÷ 6 = 120°
3. At the vertex where the shapes share a side, the angles around the point on the "outside" total 360°.
4. The interior angles of the pentagon and hexagon account for 108° + 120° = 228°.
5. Angle x = 360° - 228° = 132°

Examiner commentary: This problem requires knowledge of interior angle formulae AND understanding that angles at a point sum to 360°. Writing out the formula for each shape shows strong method. State which angle rule you use — "angles at a point add up to 360°."

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Problem 7: Area and Perimeter

Question: A rectangular garden measuring 20m by 15m has a circular pond with a diameter of 6m in the centre. Find the area of the garden not covered by the pond. Give your answer to 1 decimal place. Use π = 3.14.

Full solution:

1. Area of rectangle = 20 x 15 = 300m²
2. Radius of circle = 6 ÷ 2 = 3m
3. Area of circle = π x r² = 3.14 x 3² = 3.14 x 9 = 28.26m²
4. Area not covered by pond = 300 - 28.26 = 271.7m² (to 1 d.p.)

Examiner commentary: Remember to halve the diameter to get the radius. A very common error is using the diameter in the area formula. Also note: the question says "use π = 3.14" — if you use a different value, your answer will differ and may be marked wrong.

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Problem 8: Statistics

Question: Five friends measure their heights in cm: 142, 138, 155, 142, 148. Calculate the mean, median, and mode. Which average best represents the group? Explain.

Full solution:

1. Mean: (142 + 138 + 155 + 142 + 148) ÷ 5 = 725 ÷ 5 = 145cm
2. Median: Arrange in order: 138, 142, 142, 148, 155. Middle value = 142cm
3. Mode: 142 appears twice (the most). Mode = 142cm
4. Best average: The mean (145cm) is the best representation because it uses all five values and is not unduly affected by any outlier. The median and mode are both 142cm, which is the second-shortest height and does not represent the taller members of the group well. The mean gives a more balanced picture.

Examiner commentary: Calculating all three averages correctly earns the calculation marks. The "which is best" question is a reasoning question — you must give a mathematical reason, not just name one. Saying "because it uses all the data" is the classic justification for the mean.

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Problem 9: Measures and Money

Question: A plumber charges a £30 call-out fee plus £18.50 per hour. A job takes 2 hours 45 minutes. How much does the plumber charge?